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Applied Mathematics E-Notes, 14(2014), 107-115 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
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Existence Of Solutions For A Robin Problem
Involving The p(x)-Laplacian
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Mostafa Allaouiy
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Received 9 March 2014
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Abstract
We study the existence of weak solutions for a parametric Robin problem
driven by the p(x)-Laplacian. Our approach relies on the variable exponent theory of generalized Lebesgue-Sobolev spaces, combined with adequate variational
methods and the Mountain Pass Theorem.
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1
Introduction
The purpose of this article is to study the existence of solutions for the following
problem:
(
a(x)jujq(x) 2 u + b(x)jujr(x) 2 u ; in ;
p(x) u =
(1)
p(x) 2
jrujp(x) 2 @u
u = 0;
in @ ,
@ + (x)juj
where
RN (N
2) is a bounded smooth domain, @u
@ is the outer unit normal
derivative on @ , is a positive number, p is Lipschitz continuous on , 2 L1 (@ )
with
:= inf x2@ (x) > 0, and q; r are continuous functions on
with q :=
inf x2 q(x) > 1, r := inf x2 r(x) > 1, a(x); b(x) > 0 for x 2 such that a 2 L (x) ( ),
p (x)
(x) = p(x)p(x)q(x) and b 2 L (x) ( ), (x) = p (x)
r(x) . Here
p (x) =
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(
We will use the notations such as h
h := inf h(x)
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N p(x)
N p(x) ;
+1;
if p(x) < N;
if p(x) N:
and h+ where
h(x)
h+ := sup h(x) < +1:
x2
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x2
Throughout this paper, assuming the condition
1<q
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q+ < p
p+ < r
r+ < (p ) and p+ < N:
Mathematics Subject Classi…cations: 35J48, 35J60, 35J66.
of Applied Mathematics, University Mohamed I, Oujda, Morocco
y Department
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(2)
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Existence of Solutions for a Robin Problem
The main interest in studying such problems arises from the presence of the p(x)Laplace operator div(jrujp(x) 2 ru), which is a natural extension of the classical pLaplace operator div(jrujp 2 ru) obtained in the case when p is a positive constant.
However, such generalizations are not trivial since the p(x)- Laplace operator possesses
a more complicated structure than p Laplace operator; for example, it is inhomogeneous.
Nonlinear boundary value problems with variable exponent have received considerable attention in recent years. This is partly due to their frequent appearance in
applications such as the modeling of electro-rheological ‡uids [12, 13, 17] and image
processing [4], but these problems are very interesting from a purely mathematical
point of view as well. Many results have been obtained on this kind of problems; see
for example [1, 3, 5, 6, 8, 14, 15, 16]. In [5], the authors have studied the case a(x) = 1,
b(x) = 0 and q(x) = p(x) , they proved that the existence of in…nitely many eigenvalue
sequences. Unlike the p-Laplacian case, for a variable exponent p(x) (6= constant),
there does not exist a principal eigenvalue and the set of all eigenvalues is not closed
under some assumptions. Finally, they presented some su¢ cient conditions for the
in…mum of all eigenvalues to be zero and positive, respectively.
The main result of this paper is as follows.
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THEOREM 1. Assume p is Lipschitz continuous, q; r 2 C+ ( ) and condition (2) is
ful…lled. Then there exists
> 0 such that for any 2 (0; ), problem (1) possesses
a nontrivial weak solution.
This article is organized as follows. First, we will introduce some basic preliminary
results and lemmas in Section 2. In Section 3, we will give the proof of our main result.
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Preliminaries
For completeness, we …rst recall some facts on the variable exponent spaces Lp(x) ( )
and W 1;p(x) ( ). For more details, see [9, 10]. Suppose that
is a bounded open
domain of RN with smooth boundary @ and p 2 C+ ( ) where
C+ ( ) =
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p 2 C( ) and inf p(x) > 1 :
x2
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Denote by p := inf x2 p(x) and p+ := supx2 p(x). De…ne the variable exponent
Lebesgue space Lp(x) ( ) by
Z
p(x)
Lp(x) ( ) = u : ! R is measurable and
juj
dx < +1 ;
with the norm
jujp(x) = inf
> 0;
Z
u
p(x)
dx
1 :
De…ne the variable exponent Sobolev space W 1;p(x) ( ) by
W 1;p(x) ( ) = fu 2 Lp(x) ( ) : jruj 2 Lp(x) ( )g;
M. Allaoui
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with the norm
kuk = inf
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>0:
Z
p(x)
ru
+
u
p(x)
!
)
dx
1 ;
kuk = jrujp(x) + jujp(x) :
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(
We refer the reader to [8, 9] for the basic properties of the variable exponent
Lebesgue and Sobolev spaces.
LEMMA 1 (cf. [10]). Both (Lp(x) ( ); j jp(x) ) and (W 1;p(x) ( ); k k) are separable
and uniformly convex Banach spaces.
LEMMA 2 (cf. [10]). Hölder inequality holds, namely
Z
0
juvj dx 2 jujp(x) jvjp0 (x) for all u 2 Lp(x) ( ) and v 2 Lp (x) ( );
where
1
p(x)
+
1
p0 (x)
= 1.
h(x)
LEMMA 3 (cf. [2]). Assume that h 2 L1
2 Lp(x) ( ),
+ ( ) and p 2 C+ ( ). If juj
then we have
n
o
n
o
h
h+
h(x)
h
h+
min jujh(x)p(x) ; jujh(x)p(x)
juj
max jujh(x)p(x) ; jujh(x)p(x) :
p(x)
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LEMMA 4 (cf. [9]). Assume that
is bounded and smooth.
(i) If p is Lipschitz continuous and p+ < N , then for h 2 L1
+ ( ) with p(x)
1;p(x)
p (x) there is a continuous embedding W
( ) ,! Lh(x) ( ).
(ii) If p 2 C( ) and 1
q(x) < p (x) for x 2
(
p (x) =
N p(x)
N p(x) ;
+1;
where
if p(x) < N;
if p(x) N;
then there is a compact embedding W 1;p(x) ( ) ,! Lq(x) ( ).
Now, we introduce a norm, which will be used later. Let
inf x2@ (x) > 0 and for u 2 W 1;p(x) ( ), de…ne
(
Z
Z
p(x)
u
ru
kuk = inf
dx +
>0:
(x)
@
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h(x)
2 L1 (@ ) with
p(x)
!
d
:=
)
1 :
Then, by Theorem 2.1 in [7], k:k is also a norm on W 1;p(x) ( ) which is equivalent to
k:k.
An important role in manipulating the generalized Lebesgue-Sobolev spaces is
played by the mapping de…ned by the following.
R
R
p(x)
p(x)
LEMMA 5 (cf. [7]). Let I (u) =
jruj
dx + @ (x) juj
d with
> 0.
1;p(x)
For u 2 W
( ) we have that
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Existence of Solutions for a Robin Problem
(i) kuk < 1(= 1; > 1) , I (u) < 1(= 1; > 1),
(ii) kuk
1 ) kukp
(iii) kuk
1 ) kuk
+
p
I (u)
kukp , and
I (u)
kukp .
+
Here, problem (1) is stated in the framework of the generalized Sobolev space
X := W 1;p(x) ( ).
The Euler-Lagrange functional associated with (1) is de…ned as
: X ! R,
Z
Z
Z
(x) p(x)
a(x) q(x)
1
p(x)
jruj
dx +
juj
d
juj
dx
(u) =
p(x)
p(x)
q(x)
@
Z
b(x) r(x)
juj
dx:
r(x)
We say that u 2 X is a weak solution of (1) if
Z
Z
p(x) 2
p(x)
jruj
rurv dx +
(x) juj
@
Z
Z
q(x) 2
r(x)
=
a(x) juj
uvdx +
b(x) juj
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uvd
2
uvdx
for all v 2 X.
Standard arguments imply that
2 C 1 (X; R) and
Z
Z
p(x) 2
p(x)
h 0 (u); vi =
jruj
rurv dx +
(x) juj
@
Z
Z
q(x) 2
r(x)
a(x) juj
uvdx
b(x) juj
2
uvd
2
uvdx;
for all u; v 2 X. Thus the weak solutions of (1) coincide with the critical points of
. If such a weak solution exists and is nontrivial, then the corresponding is an
eigenvalue of problem (1).
Next, we write 0 as
0
=A
B;
where A; B : X ! X 0 are de…ned by
Z
Z
p(x) 2
hA(u); vi =
jruj
rurv dx +
@
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2
and
hB(u); vi =
Z
a(x) juj
q(x) 2
uv dx +
Z
(x) juj
p(x) 2
r(x) 2
b(x) juj
uv d
uv dx:
Denote by M; C; Ci ; i = 1; 2::: the general positive constants which are the exact
values may change from line to line.
LEMMA 6 (cf. [11]). A satis…es condition (S + ), namely, un * u, in X and
lim suphA(un ); un ui 0, imply un ! u in X.
REMARK 1. Noting that 0 is still of type (S + ). Hence, any bounded (PS)
sequence of
in the re‡exive Banach space X has a convergent subsequence.
M. Allaoui
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Proof of Main Result
For the proof of our theorem, we will use the Mountain Pass Lemma. We need to
establish some lemmas.
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LEMMA 7. The functional
satis…es the Palais-Smale condition (PS).
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PROOF. Suppose that (un )
X is a (PS) sequence; that is,
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sup j
(un )j
M (for any n or as n ! 1 ?) and
0
(un ) ! 0 as n ! 1:
Let us show that (un ) is bounded in X. Assume kun k > 1 for convenience. Since
(un ) is bounded, we have for n large enough:
M +1
1 0
(un ); un i +
h (un ); un i
Z
Z r
Z
1
(x)
a(x)
p(x)
p(x)
=
jrun j
dx +
jun j
d
jun jq(x) dx
p(x)
q(x)
@ p(x)
Z
Z
Z
1
b(x)
p(x)
jrun j
dx +
(x)jun jp(x) d
jun jr(x) dx
r(x)
r
@
Z
Z
1 0
r(x)
q(x)
+
a(x)jun j
dx +
b(x) jun j
dx +
h (un ); un i
r
r
r
Z
Z
1
1
I (un )
a(x)jun jq(x) dx
b(x)jun jr(x) dx
I (un )
p+
q
r
r
Z
Z
1 0
a(x)jun jq(x) dx +
b(x)jun jr(x) dx +
h (un ); un i
+
r
r
r
1
1
1
1
q+
I (un )
C1 jaj (x) kun k
+
p
r
q
r
1
0
(un ) X 0 kun k
r
1
1
1
1
C2
p
q+
kun k
C1 jaj (x) kun k
kun k
+
p
r
q
r
r
+
1
1
1
1
C2
p
kun k;
C3 kun k
C1 jaj (x) kun kq
+
p
r
q
r
r
(un )
1
h
r
0
hence (un ) is bounded in X since q
q+ < p
p+ < r . The proof is completed.
LEMMA 8. There exists
> 0 such that for any 2 (0;
such that
(u)
> 0 for any u 2 X with kuk = .
) there exist ;
>0
PROOF. Using Lemma 4, there exists a positive constant C4 such that
jujp(x)
C4 kuk
and jujp
(x)
C4 kuk
for all u 2 X:
(3)
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Existence of Solutions for a Robin Problem
Fix 2]0; 1[ such that < C14 . Then relation (3) implies jujp(x) < 1, jujp
all u 2 X with kuk = . Using Lemmas 2 and 3, we obtain
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and
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Z
a(x) juj
Z
b(x)jujr(x) dx
q(x)
and
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(u)
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j ujr(x)
(x)
2 jaj
p(x)
q(x)
2jbj
p (x)
r(x)
Z
a(x)jujq(x) dx
2jaj
q
(x) C4
kuk
Z
b(x)jujr(x) dx
2jbj
r
(x) C4
kuk
1
p+
Z
Z
p(x)
jruj
q
(x)
jujp(x) ;
r
(x)
jujp
(x)
(4)
;
(5)
q
;
(6)
r
;
(7)
dx +
Z
(x)jujp(x) d
Z
q
@
q(x)
a(x) juj
dx
r(x)
r
2jbj
r
(x) C4
r
kuk
:
Putting
= min
(
q
8C4q
p+ q
+
r p r
;
p+ jaj (x) 8C4r p+ jbj
for any u 2 X with kuk = , there exists
=
(u)
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juj
b(x) juj
dx
r
+
1
q
kukp
2jaj (x) C4q kuk
p+
q
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2 jbj
(x)
for all u 2 X with kuk = . Hence, from (6), (7) we deduce that for any u 2 X with
kuk = , we have
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q(x)
2jaj
< 1, for
for all u 2 X with kuk = . Combining (3), (4) and (5), we obtain
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dx
(x)
p+
(x)
)
(8)
=(2p+ ) such that
> 0:
This completes the proof.
LEMMA 9. There exists
small enough.
2 X such that
0,
6= 0 and
(t ) < 0, for t > 0
M. Allaoui
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PROOF. Let 2 C01 ( ),
0, 6= 0 and t 2 (0; 1). We have
Z p(x)
Z
Z
t
tp(x) (x) p(x)
tq(x) q(x)
p(x)
(t ) =
jr j
dx +
j j
d
j j
dx
a(x)
p(x)
p(x)
q(x)
@
Z
tr(x) r(x)
j j
dx
b(x)
r(x)
Z
Z
+ Z
tp
tq
p(x)
p(x)
q(x)
jr j
dx +
(x) j j
d
a(x) j j
dx
+
p
q
@
+ Z
tr
r(x)
b(x) j j
dx
r+
Z
Z
tp
p(x)
p(x)
jr j
dx +
(x) j j
d
p
@
Z
Z
+
tq
r(x)
q(x)
b(x) j j
dx :
a(x)
j
j
dx
+
q+
1
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Then, for any t <
0<
we conclude that
, with
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9
R
R
q(x)
r(x)
<
p
a(x) j j
dx + b(x) j j
dx =
< min 1;
;
R
R
p(x)
p(x)
:
;
q+
jr j
dx + @ (x) j j
d
p
(t ) < 0:
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q+
The proof is complete.
We now turn to the proof of Theorem 1. To apply the Mountain Pass Theorem, we
need to prove that
(tu) ! 1 as t ! +1;
for a certain u 2 X. Let ! 2 C01 ( ), ! 0, ! 6= 0 and t > 1. We have
Z
Z
Z p(x)
tp(x) (x)
tq(x)
t
p(x)
p(x)
q(x)
jr!j
dx +
j!j
d
a(x)
j!j
dx
(t!) =
p(x)
p(x)
q(x)
@
Z
tr(x) r(x)
b(x)
j!j
dx
r(x)
Z
Z
Z
+
tp
tq
p(x)
p(x)
q(x)
jr!j
dx +
(x) j!j
d
a(x) j!j
dx
+
p
q
@
Z
tr
r(x)
b(x) j!j
dx:
r+
Since q ; p+ < r we have (t!) ! 1 as t ! +1: It follows that there exists e 2 X
such that kek > and
(e) < 0. According to the Mountain Pass Theorem,
admits a critical value
which is characterized by
= inf sup
g2 t2[0;1]
(g(t));
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where
= fg 2 C([0; 1]; X) : g(0) = 0 and g(1) = eg:
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Existence of Solutions for a Robin Problem
This completes the proof.
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Acknowledgment. The author thanks the referees for their careful reading of the
manuscript and insightful comments.
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